Question: Solve for $q$, $ -\dfrac{8}{5q - 3} = \dfrac{3q + 6}{20q - 12} + \dfrac{2}{10q - 6} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5q - 3$ $20q - 12$ and $10q - 6$ The common denominator is $20q - 12$ To get $20q - 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{5q - 3} \times \dfrac{4}{4} = -\dfrac{32}{20q - 12} $ The denominator of the second term is already $20q - 12$ , so we don't need to change it. To get $20q - 12$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{2}{10q - 6} \times \dfrac{2}{2} = \dfrac{4}{20q - 12} $ This give us: $ -\dfrac{32}{20q - 12} = \dfrac{3q + 6}{20q - 12} + \dfrac{4}{20q - 12} $ If we multiply both sides of the equation by $20q - 12$ , we get: $ -32 = 3q + 6 + 4$ $ -32 = 3q + 10$ $ -42 = 3q $ $ q = -14$